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Chapter 13: Problem 63

Express each sum using summation notation. \(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\cdots+(-1)^{6}\left(\frac{1}{3^{6}}\right)\)

Short Answer

Expert verified
\(\sum_{n=0}^{6} (-1)^n \left(\frac{1}{3^n}\right)\)

Step by step solution

01

Identify the Pattern

Observe the given series: \(1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \cdots + (-1)^{6}\left(\frac{1}{3^{6}}\right)\). Notice that this is an alternating series where each term involves a power of \(\frac{1}{3}\).
02

Write the General Term

The general term of the series can be written as \((-1)^{n}\left(\frac{1}{3^{n}}\right)\). Here, each term鈥檚 sign alternates due to \((-1)^{n}\) and the denominator is a power of 3.
03

Define the Range of the Summation

Identify the range of the summation. The given series starts at \(n=0\) and ends at \(n=6\). This means the summation is over \(n\) going from 0 to 6.
04

Combine into Summation Notation

Combine the identified components into summation notation to express the sum: \[\sum_{n=0}^{6} (-1)^n \left(\frac{1}{3^n}\right)\]}],

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Series
An alternating series is a series where the terms alternate in sign. In other words, the signs change from positive to negative or vice versa sequentially. This can be observed in the given series:
  • First term: 1
  • Second term: -1/3
  • Third term: 1/9
  • Fourth term: -1/27
Notice how the sign changes in each term. The presence of \((-1)^n \) in the general term helps in alternating the signs of the series. When \((-1)^n \) is raised to an even power \(n = 0, 2, 4, 6, \), the term is positive. When it is raised to an odd power \(n = 1, 3, 5\), the term is negative.
General Term
The general term of a series is a formula that represents the terms in the series for a given value of the index of summation. In the given exercise, the general term is represented as \((-1)^n \left(\frac{1}{3^n} \right)\).
  • \((-1)^n \) alternates the sign of the terms.
  • \(\frac{1}{3^n} \) indicates the sequence follows powers of 3 in the denominator.
For example, when \(n=0\), the term is \(1\). When \(n=1\), the term is \(-\frac{1}{3}\). This pattern continues with \((-1)^n\) alternating the sign and \(\frac{1}{3^n}\) representing the fractions involved.
Range of Summation
The range of summation tells us the starting and ending values of the index for which the summation is considered. For the given series, the index of summation is represented by \(n\) and it ranges from 0 to 6.

This means the summation notation will include values of \(n\) starting from 0, going up to and including 6. \[\textstyle \sum \limits \_{n=0}^{6} \left((-1)^n \frac{1}{3^n} \right)\]Here, \(n=0\) gives the first term, \(n=1\) gives the second term, and so on, until \(n=6\). The formula adds up each term in this range, considering the alternating sign and denominator pattern defined by the general term.

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Most popular questions from this chapter

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1 \cdot 2+3 \cdot 4+5 \cdot 6+\cdots+(2 n-1)(2 n)=\frac{1}{3} n(n+1)(4 n-1) $$

Challenge Problem Paper Creases If a sheet of paper is folded in half by folding the top edge down to the bottom edge, one crease will result. If the folded paper is folded in the same manner, the result is three creases. With each fold, the number of creases can be defined recursively by \(c_{1}=1, c_{n+1}=2 c_{n}+1\) (a) Find the number of creases for \(n=3\) and \(n=4\) folds. (b) Use the given information and your results from part (a) to find a formula for the number of creases after \(n\) folds, \(c_{n}\), in terms of the number of folds alone. (c) Use the Principle of Mathematical Induction to prove that the formula found in part (b) is correct for all natural numbers. (d) Tosa Tengujo is reportedly the world's thinnest paper with a thickness of \(0.02 \mathrm{~mm}\). If a piece of this paper could be folded 25 times, how tall would the stack be?

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+2+2^{2}+\cdots+2^{n-1}=2^{n}-1 $$

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ -2-3-4-\cdots-(n+1)=-\frac{1}{2} n(n+3) $$

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve the system: \(\left\\{\begin{array}{l}4 x+3 y=-7 \\ 2 x-5 y=16\end{array}\right.\)
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